How much kerosene must be burned in order to turn 400 g of water taken at the boiling point into steam?

Initial data: m1 (mass of water) = 400 g = 0.4 kg; water is taken at boiling point (t = 100 ºС).

Reference data: L (specific heat of vaporization of water) = 2.3 * 10 ^ 6 J / kg; q (specific heat of combustion of kerosene) = 46 MJ / kg = 46 * 10 ^ 6 J / kg.

The required amount of kerosene is determined from the equality: L * m1 = q * m2, whence m2 = L * m1 / m2.

Let’s calculate: m2 = 2.3 * 10 ^ 6 * 0.4 / (46 * 10 ^ 6) = 2.3 * 0.4 / 46 = 0.02 kg or 20 g.

Answer: It is necessary to burn 20 grams of kerosene.